A particle moves in one dimension such that its velocity as a function of time is given by v(t) = t - 2 , where v is in m/s and t is in seconds. The ratio of the total distance covered to the net displacement of the particle during the time interval from t = 0 to t = 6 s is:

Step-by-step solution

The velocity of the particle is v(t) = t - 2 . The particle comes to rest when v(t) = 0 , which gives t = 2 s . For 0 t For 2 The net displacement S is given by the integral of velocity: S = ₀⁶ (t - 2) dt = [ t^2 2 - 2t ]₀⁶ = ( 36 2 - 12 ) - 0 = 18 - 12 = 6 m The total distance D is the integral of the speed |v(t)| : D = ₀² -(t - 2) dt + ₂⁶ (t - 2) dt D = [ 2t - t^2 2 ]₀² + [ t^2 2 - 2t ]₂⁶ D = (4 - 2) + (18 - 12) - (2 - 4) = 2 + 6 - (-2) = 10 m The ratio of the total distance to the net displacement is: D S = 10 6